Let $\mathcal{P}$ be a convex lattice polygon with $n$ vertices and let $\mathcal{L}$ be the set of all lattice points inside $\mathcal{P}$. For every $n \geq 5$, does there exist a point in $\mathcal{L}$ such that it also lies in the convex polygon bounded by (all) the diagonals of $\mathcal{P}$? How many such points are there? (//By diagonals I mean of course the lines different from the sidelines of the polygon which are connecting two vertices of $\mathcal{P}$.)

I proved a while ago that for $n=5$ there is such a point in $\mathcal{L}$. I also managed to show this now for $n \geq 6$ using a similar argument, yet it got more involved and I still need to check for potential bugs. Any ideas for the general case?

I suspect there are simple counterexamples for n=6, so I may be misunderstanding something. Can you say more about what interior region is supposed to have a lattice point? Gerhard "Ask Me About System Design" Paseman, 2011.11.17
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Gerhard PasemanNov 17 '11 at 10:07

What do you mean by "the convex pentagon bounded by (all) the diagonals of P"? usually the do not bound a pentagon.
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Fedor PetrovNov 17 '11 at 11:41

I am still having difficulty understanding the phrase, "the convex polygon bounded by all the diagonals of $P$." In general, there is no convex polygon bounded by all the diagonals, if by "bounded" you mean, "forming the boundary of." There are many convex polygons, each bounded by a subset of the diagonals...
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Joseph O'RourkeNov 17 '11 at 17:56

1 Answer
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For odd $n>5$, one could consider the polygon bounded by the longest diagonals.
It may be defined as the intersection of the half-planes containing $(n+1)/2$ vertices of $P$.
For $n=9$, this intersection may be empty (for example, if the nine vertices form three triples and each of the triples is placed very close to a vertex of a regular triangle).
For $n=7$, this intersection is non-empty. However, it may be free of lattice points:
take, for example, the polygon $P$ with vertices
$[0,1],
[1,0],
[2,0],
[3,2],
[3,3],
[1,3],
[0,2]$.